use*_*499 11 python-2.7 pandas scikit-learn
我有两个数据帧 - df1和df2.
df1 has row1,row2,row3,row4,row5
df2 has row2,row5
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我希望有一个新的数据帧df1-df2.也就是说,结果数据帧的行应为 - row1,row3,row4.
您可以使用该index.difference()功能
import numpy as np
import pandas as pd
df1 = pd.DataFrame(np.random.randn(5, 2), index= ['row' + str(i) for i in range(1, 6)])
df1
0 1
row1 0.249451 -0.107651
row2 1.295390 -1.773707
row3 -0.893647 -0.683306
row4 -1.090551 0.016833
row5 0.864612 0.369138
df2 = pd.DataFrame(np.random.randn(2, 2), index= ['row' + str(i) for i in [2, 5]])
df2
0 1
row2 0.549396 -0.675574
row5 1.348785 0.942216
df1.loc[df1.index.difference(df2.index), ]
0 1
row1 0.249451 -0.107651
row3 -0.893647 -0.683306
row4 -1.090551 0.016833
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您可以使用pandas.concat逐行连接两个数据帧,然后drop_duplicates删除其中的所有重复行.
In [1]: import pandas as pd
df_1 = pd.DataFrame({"A":["foo", "foo", "foo", "bar"], "B":[0,1,1,1], "C":["A","A","B","A"]})
df_2 = pd.DataFrame({"A":["foo", "bar", "foo", "bar"], "B":[1,0,1,0], "C":["A","B","A","B"]})
In [2]: df = pd.concat([df_1, df_2])
In [3]: df
Out[3]:
A B C
0 foo 0 A
1 foo 1 A
2 foo 1 B
3 bar 1 A
0 foo 1 A
1 bar 0 B
2 foo 1 A
3 bar 0 B
In [4]: df.drop_duplicates(keep=False)
Out[4]:
A B C
0 foo 0 A
2 foo 1 B
3 bar 1 A
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df = df1.drop_duplicates().merge(df2.drop_duplicates(), on=df2.columns.to_list(),
how='left', indicator=True)
df.loc[df._merge=='left_only',df.columns!='_merge']
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请注意,删除重复项用于最大限度地减少比较。没有他们它也能工作。
最好的方法是比较行内容本身,而不是比较索引或一/两列,并且相同的代码也可用于其他过滤器(例如“both”和“right_only”)以获得类似的结果。index.difference仅适用于基于唯一索引的比较
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